1 / 29

“Rate-Optimal” Resource-Constrained Software Pipelining

“Rate-Optimal” Resource-Constrained Software Pipelining. Objectives. Establish good bounds Help compiler writers Help architects Study pragmatic issues when implemented as an option of compilers Study its payoffs. A Motivating Example. for i = 0 to n do

calais
Télécharger la présentation

“Rate-Optimal” Resource-Constrained Software Pipelining

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. “Rate-Optimal” Resource-Constrained Software Pipelining \course\cpeg421-08s\Topic-7a.ppt

  2. Objectives • Establish good bounds • Help compiler writers • Help architects • Study pragmatic issues when implemented as an option of compilers • Study its payoffs \course\cpeg421-08s\Topic-7a.ppt

  3. A Motivating Example for i = 0 to n do 0: a [i] = X + d [i - 2]; 1: b [i] = a [i] * F + f [i - 2] + e [i - 2]; 2: c [i] = Y - b [i]; 3: d [i] = 2 * c [i]; 4: e [i] = X - b [i]; 5: f [i] = Y + b [i] end \course\cpeg421-08s\Topic-7a.ppt

  4. 2 L: for ( i = 0; i < n; i ++) { 3 0 S : a [i] = X + d [i - 2]; 0 S : b [i] = a [i] * F + f [i - 2] + e [i - 2]; 1 1 2 S : c [i] = Y - b [i]; 2 S : d [i] = 2 * c [i]; 2 2 3 S : e [i] = X - b [i]; 4 4 5 S : f [i] = Y + b [i] 5 } Data Dependence Graph Program Representation Assume all statements have a delay = 1 \course\cpeg421-08s\Topic-7a.ppt

  5. Question • How fast can L run (I.e.optimal computation rate) without resource constraints? • How many FUs it needs minimally to achieve the optimal rate? \course\cpeg421-08s\Topic-7a.ppt

  6. Schedule A: t (i, Sj) = 2i + tsj where: ts0 = 0 ts1 = 1 ts2 = ts4 = ts5 = 2 ts3 = 3 iter #1 #2 #3 . . . 0 S 0 1 S 1 2 S , S , S S 2 4 5 0 3 S S 3 1 4 S , S , S S 2 4 5 0 5 S S 3 1 6 S , S , S 2 4 5 7 S . . . 3 Schedule A : # of FUs = 4 \course\cpeg421-08s\Topic-7a.ppt

  7. Can we do better? \course\cpeg421-08s\Topic-7a.ppt

  8. Schedule B : t (i, Sj) = 2i + tsj where: ts0 = 0 ts1 = 1 ts2 = ts4 = 2 ts3 = ts5 = 3 iter #1 #2 #3 . . . 0 S 0 1 S 1 2 S , S S 2 4 0 3 S S S 3, 5 1 4 S , S S 2 4 0 5 S S S 3, 5 1 6 S , S 2 4 7 S S . . . 3, 5 Schedule B : # of FUs = 3 \course\cpeg421-08s\Topic-7a.ppt

  9. Problem Statements Assumptions: homogeneous function units Problem 0: Given a loop L, determine a “rate-optimal” schedule for L which uses minimum # of FUs. Problem I (FIxed Rate SofTware Pipelining with minimum Resource - FIRST): Given a loop L and a fixed initiation rate determine a schedule for L which uses minimum # of FUs. Problem II (REsource Constrained SofTware Pipelining - REST) Given a loop L and the # of Fus, determine an optimal schedule which can run under the given resource constraints. \course\cpeg421-08s\Topic-7a.ppt

  10. Problem Formulation of FIRST How to formulate resource constraints into a linear form? \course\cpeg421-08s\Topic-7a.ppt

  11. R = Max (width) Min R • Subject to: • all dependence constraints • R is the maximum width • for a fixed rate (or period) The SWP kernel: --- The “frustum” \course\cpeg421-08s\Topic-7a.ppt

  12. i time 0 1 2 . . . N-1 r 0 1 1 1 2 1 3 1 . . . II-1 1 Min (max ari) R Frustum A Contribution of node i to step r’s resource requirement Now R = max ari r [ 0, II - 1] \course\cpeg421-08s\Topic-7a.ppt

  13. FIRST: A Linear Programming Formulation Example 0 1 2 3 4 5 a00 a01 a02 a03 a04 a05 a10 a11 a12 a13 a14 a15 So for node I: ti = ki II + (aoi , a1i ) * 0 1 Where ari = 1 means node i start at step r = 0 otherwise \course\cpeg421-08s\Topic-7a.ppt

  14. FIRST: A Linear Programming Formulation Minimize R Subject to R: Max T T = T Periodic Dependence Constraints and are integers \course\cpeg421-08s\Topic-7a.ppt

  15. Example Revisited Min R Subjected to: - + + + + + ³ R ( a a a a a a ) 0 00 01 02 03 04 05 - + + + + + ³ R ( a a a a a a ) 0 10 11 12 13 14 15 × + × + × = 2 k 0 a 1 a t 0 00 10 0 × + × + × = 2 k 1 0 a 1 a t 1 01 11 . . × + × + × = 2 k 0 a 1 a t 5 05 15 5 + = a a 1 00 10 + = a a 1 01 11 . . + = a a 1 05 15 - . . t t ³ 1 0 1 . . - t t ³ - 1 4 3 - t t ³ - 1 5 3 . . ³ ³ ³ t 0 , k 0 , a 0 1 i ri \course\cpeg421-08s\Topic-7a.ppt

  16. Solution Schedule C t(i, s) = 2i + ts where t0 = 0 t1 = 1 t2 = 2 t3 = 3 t4 = 3 t5 =2 so, # of FUs = 3! \course\cpeg421-08s\Topic-7a.ppt

  17. Linear Programming • A Linear Program is a problem that can be expressed in the following form: • minimize cx • subject to • Ax = b • x >= 0 • where x: vector of variables to be solved • A: matrix of known coefficients • c, b: vectors of known coefficients • cx: it is called the objective function • Ax=b is called the constraints \course\cpeg421-08s\Topic-7a.ppt

  18. Linear Programming • “programming” actually means “planning” here. • Importance of LP • many applications • the existence of good general-purpose techniques for finding optimal solutions \course\cpeg421-08s\Topic-7a.ppt

  19. Integer Linear Programming(ILP) • Integer programming have proved valuable for modeling many and diverse types of problems in : • planning, • routing, • scheduling, • assignment, and • design. • Industry applications: • transportation, energy, telecommunications, and manufacturing \course\cpeg421-08s\Topic-7a.ppt

  20. Integer Linear Programming • Classification: • mixed integer (part of variables) • pure integer (all variables) • zero-one (only takes 0 or 1) • Much harder to solve • Many existing tools and product • solve the LP/ILP problem and get optimal solution • help to model the problem, and then solve the problem \course\cpeg421-08s\Topic-7a.ppt

  21. Quiz How to handle cases where di 1 for node i ? >= \course\cpeg421-08s\Topic-7a.ppt

  22. The “Trick” At time step r, the # of FUs required Contributed by node i {Started at steps in [0, (t + di-1)%II]} a((r - 1)%II)i Note: if actor i requires a FU at time r or 0 otherwise. So objective function becomes: Max Min \course\cpeg421-08s\Topic-7a.ppt

  23. How to extend the FIRST formulation to heterogeneous function units? \course\cpeg421-08s\Topic-7a.ppt

  24. Mk - 0 FU(i) = k Heterogeneous Function Units Solution Hints: “weighted Sum”! For FU of type k: Mk = max " Î - r [ 0 , II 1 ] and so, the objective should be min Ck Mk Where h is the total # of FU types " Î - " Î - k [ 0 , h 1 ], r [ 0 , II 1 ] \course\cpeg421-08s\Topic-7a.ppt

  25. Solution of the REST Resource - constraint rate-optimal software pipelining problem \course\cpeg421-08s\Topic-7a.ppt

  26. Hints Based on scheme for “FIRST” and play “trial-and improve” \course\cpeg421-08s\Topic-7a.ppt

  27. Solution Space for REST MII = max { RecMII, ResMII} 1 2 \course\cpeg421-08s\Topic-7a.ppt

  28. Max cycles C d(C) m(C) RecMII = # of nodes # of FUs ResMII = MII = max {RecMII, ResMII } Note: The bounds of initiation interval 1. Loop-carried dependence constraints: 2. Resource constraints As a result: \course\cpeg421-08s\Topic-7a.ppt

  29. Other Work • Extend the technique to “more benchmarks” and establish “bounds” • Extend the framework to include register constraints • Extend the model for pipelined architectures (with structural “hazards”) • Extend the model to consider superscalar and multithreaded with dynamic scheduling support. \course\cpeg421-08s\Topic-7a.ppt

More Related